Problem: What is the average value of $14-6x^2$ on the interval $[-1,3]$ ?
Solution: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=14-6x^2}$, ${a=-1}$ and ${b=3}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{-1}}^{ {3}} ({14-6x^2})\,dx}{{3}-{(-1)}} \\\\ &=\dfrac{\Big[14x-2x^3\Big]_{-1}^{3}}{4} \\\\ &=\dfrac{-12-(-12)}{4} \\\\ &=0 \end{aligned}$ In conclusion, the average value of $14-6x^2$ on the interval $[-1,3]$ is $0$.